- Practice Problems
- Assignment Problems
- Show all Solutions/Steps/ etc.
- Hide all Solutions/Steps/ etc.
- Solutions and Solution Sets
- Applications of Linear Equations
- Preliminaries
- Graphing and Functions
- Calculus II
- Calculus III
- Differential Equations
- Algebra & Trig Review
- Common Math Errors
- Complex Number Primer
- How To Study Math
- Cheat Sheets & Tables
- MathJax Help and Configuration
- Notes Downloads
- Complete Book
- Practice Problems Downloads
- Complete Book - Problems Only
- Complete Book - Solutions
- Assignment Problems Downloads
- Other Items
- Get URL's for Download Items
- Print Page in Current Form (Default)
- Show all Solutions/Steps and Print Page
- Hide all Solutions/Steps and Print Page
Section 2.2 : Linear Equations
Solve each of the following equations and check your answer.
- \(4x - 7\left( {2 - x} \right) = 3x + 2\) Solution
- \(2\left( {w + 3} \right) - 10 = 6\left( {32 - 3w} \right)\) Solution
- \(\displaystyle \frac{{4 - 2z}}{3} = \frac{3}{4} - \frac{{5z}}{6}\) Solution
- \(\displaystyle \frac{{4t}}{{{t^2} - 25}} = \frac{1}{{5 - t}}\) Solution
- \(\displaystyle \frac{{3y + 4}}{{y - 1}} = 2 + \frac{7}{{y - 1}}\) Solution
- \(\displaystyle \frac{{5x}}{{3x - 3}} + \frac{6}{{x + 2}} = \frac{5}{3}\) Solution
Word Problems on Linear Equations
Worked-out word problems on linear equations with solutions explained step-by-step in different types of examples.
There are several problems which involve relations among known and unknown numbers and can be put in the form of equations. The equations are generally stated in words and it is for this reason we refer to these problems as word problems. With the help of equations in one variable, we have already practiced equations to solve some real life problems.
Steps involved in solving a linear equation word problem: ● Read the problem carefully and note what is given and what is required and what is given. ● Denote the unknown by the variables as x, y, ……. ● Translate the problem to the language of mathematics or mathematical statements. ● Form the linear equation in one variable using the conditions given in the problems. ● Solve the equation for the unknown. ● Verify to be sure whether the answer satisfies the conditions of the problem.
Step-by-step application of linear equations to solve practical word problems:
1. The sum of two numbers is 25. One of the numbers exceeds the other by 9. Find the numbers.
Solution: Then the other number = x + 9 Let the number be x. Sum of two numbers = 25 According to question, x + x + 9 = 25 ⇒ 2x + 9 = 25 ⇒ 2x = 25 - 9 (transposing 9 to the R.H.S changes to -9) ⇒ 2x = 16 ⇒ 2x/2 = 16/2 (divide by 2 on both the sides) ⇒ x = 8 Therefore, x + 9 = 8 + 9 = 17 Therefore, the two numbers are 8 and 17.
2.The difference between the two numbers is 48. The ratio of the two numbers is 7:3. What are the two numbers? Solution: Let the common ratio be x. Let the common ratio be x. Their difference = 48 According to the question, 7x - 3x = 48 ⇒ 4x = 48 ⇒ x = 48/4 ⇒ x = 12 Therefore, 7x = 7 × 12 = 84 3x = 3 × 12 = 36 Therefore, the two numbers are 84 and 36.
3. The length of a rectangle is twice its breadth. If the perimeter is 72 metre, find the length and breadth of the rectangle. Solution: Let the breadth of the rectangle be x, Then the length of the rectangle = 2x Perimeter of the rectangle = 72 Therefore, according to the question 2(x + 2x) = 72 ⇒ 2 × 3x = 72 ⇒ 6x = 72 ⇒ x = 72/6 ⇒ x = 12 We know, length of the rectangle = 2x = 2 × 12 = 24 Therefore, length of the rectangle is 24 m and breadth of the rectangle is 12 m.
4. Aaron is 5 years younger than Ron. Four years later, Ron will be twice as old as Aaron. Find their present ages.
Solution: Let Ron’s present age be x. Then Aaron’s present age = x - 5 After 4 years Ron’s age = x + 4, Aaron’s age x - 5 + 4. According to the question; Ron will be twice as old as Aaron. Therefore, x + 4 = 2(x - 5 + 4) ⇒ x + 4 = 2(x - 1) ⇒ x + 4 = 2x - 2 ⇒ x + 4 = 2x - 2 ⇒ x - 2x = -2 - 4 ⇒ -x = -6 ⇒ x = 6 Therefore, Aaron’s present age = x - 5 = 6 - 5 = 1 Therefore, present age of Ron = 6 years and present age of Aaron = 1 year.
5. A number is divided into two parts, such that one part is 10 more than the other. If the two parts are in the ratio 5 : 3, find the number and the two parts. Solution: Let one part of the number be x Then the other part of the number = x + 10 The ratio of the two numbers is 5 : 3 Therefore, (x + 10)/x = 5/3 ⇒ 3(x + 10) = 5x ⇒ 3x + 30 = 5x ⇒ 30 = 5x - 3x ⇒ 30 = 2x ⇒ x = 30/2 ⇒ x = 15 Therefore, x + 10 = 15 + 10 = 25 Therefore, the number = 25 + 15 = 40 The two parts are 15 and 25.
More solved examples with detailed explanation on the word problems on linear equations.
6. Robert’s father is 4 times as old as Robert. After 5 years, father will be three times as old as Robert. Find their present ages. Solution: Let Robert’s age be x years. Then Robert’s father’s age = 4x After 5 years, Robert’s age = x + 5 Father’s age = 4x + 5 According to the question, 4x + 5 = 3(x + 5) ⇒ 4x + 5 = 3x + 15 ⇒ 4x - 3x = 15 - 5 ⇒ x = 10 ⇒ 4x = 4 × 10 = 40 Robert’s present age is 10 years and that of his father’s age = 40 years.
7. The sum of two consecutive multiples of 5 is 55. Find these multiples. Solution: Let the first multiple of 5 be x. Then the other multiple of 5 will be x + 5 and their sum = 55 Therefore, x + x + 5 = 55 ⇒ 2x + 5 = 55 ⇒ 2x = 55 - 5 ⇒ 2x = 50 ⇒ x = 50/2 ⇒ x = 25 Therefore, the multiples of 5, i.e., x + 5 = 25 + 5 = 30 Therefore, the two consecutive multiples of 5 whose sum is 55 are 25 and 30.
8. The difference in the measures of two complementary angles is 12°. Find the measure of the angles. Solution: Let the angle be x. Complement of x = 90 - x Given their difference = 12° Therefore, (90 - x) - x = 12° ⇒ 90 - 2x = 12 ⇒ -2x = 12 - 90 ⇒ -2x = -78 ⇒ 2x/2 = 78/2 ⇒ x = 39 Therefore, 90 - x = 90 - 39 = 51 Therefore, the two complementary angles are 39° and 51°
9. The cost of two tables and three chairs is $705. If the table costs $40 more than the chair, find the cost of the table and the chair. Solution: The table cost $ 40 more than the chair. Let us assume the cost of the chair to be x. Then the cost of the table = $ 40 + x The cost of 3 chairs = 3 × x = 3x and the cost of 2 tables 2(40 + x) Total cost of 2 tables and 3 chairs = $705 Therefore, 2(40 + x) + 3x = 705 80 + 2x + 3x = 705 80 + 5x = 705 5x = 705 - 80 5x = 625/5 x = 125 and 40 + x = 40 + 125 = 165 Therefore, the cost of each chair is $125 and that of each table is $165.
10. If 3/5 ᵗʰ of a number is 4 more than 1/2 the number, then what is the number? Solution: Let the number be x, then 3/5 ᵗʰ of the number = 3x/5 Also, 1/2 of the number = x/2 According to the question, 3/5 ᵗʰ of the number is 4 more than 1/2 of the number. ⇒ 3x/5 - x/2 = 4 ⇒ (6x - 5x)/10 = 4 ⇒ x/10 = 4 ⇒ x = 40 The required number is 40.
Try to follow the methods of solving word problems on linear equations and then observe the detailed instruction on the application of equations to solve the problems.
● Equations
What is an Equation?
What is a Linear Equation?
How to Solve Linear Equations?
Solving Linear Equations
Problems on Linear Equations in One Variable
Word Problems on Linear Equations in One Variable
Practice Test on Linear Equations
Practice Test on Word Problems on Linear Equations
● Equations - Worksheets
Worksheet on Linear Equations
Worksheet on Word Problems on Linear Equation
7th Grade Math Problems 8th Grade Math Practice From Word Problems on Linear Equations to HOME PAGE
New! Comments
Didn't find what you were looking for? Or want to know more information about Math Only Math . Use this Google Search to find what you need.
- Preschool Activities
- Kindergarten Math
- 1st Grade Math
- 2nd Grade Math
- 3rd Grade Math
- 4th Grade Math
- 5th Grade Math
- 6th Grade Math
- 7th Grade Math
- 8th Grade Math
- 9th Grade Math
- 10th Grade Math
- 11 & 12 Grade Math
- Concepts of Sets
- Probability
- Boolean Algebra
- Math Coloring Pages
- Multiplication Table
- Cool Maths Games
- Math Flash Cards
- Online Math Quiz
- Math Puzzles
- Binary System
- Math Dictionary
- Conversion Chart
- Homework Sheets
- Math Problem Ans
- Free Math Answers
- Printable Math Sheet
- Funny Math Answers
- Employment Test
- Math Patterns
- Link Partners
© and ™ math-only-math.com. All Rights Reserved. 2010 - 2022.
If you're seeing this message, it means we're having trouble loading external resources on our website.
If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.
To log in and use all the features of Khan Academy, please enable JavaScript in your browser.
Unit 6: Lesson 3
- Solving linear equations and linear inequalities | Lesson
- Understanding linear relationships | Lesson
- Linear inequality word problems | Lesson
- Graphing linear equations | Lesson
- Systems of linear inequalities word problems | Lesson
- Solving systems of linear equations | Lesson
Systems of linear equations word problems | Lesson
What are systems of linear equations word problems, and how frequently do they appear on the test.
- Understanding linear relationships
- Solving systems of linear equations
How do I solve systems of linear equations word problems?
Systems of equations examples, how do i write systems of linear equations.
- Select variables to represent the unknown quantities.
- Using the given information, write a system of two linear equations relating the two variables.
- Solve the system of linear equations using either substitution or elimination.
Let's look at an example!
- (Choice A) c − 1 = b 4 b + 6 c = 31 \begin{aligned} c-1 &= b \\ \\ 4b+6c &=31 \end{aligned} c − 1 4 b + 6 c = b = 3 1 A c − 1 = b 4 b + 6 c = 31 \begin{aligned} c-1 &= b \\ \\ 4b+6c &=31 \end{aligned} c − 1 4 b + 6 c = b = 3 1
- (Choice B) c + 1 = b 4 b + 6 c = 31 \begin{aligned} c+1 &= b \\ \\ 4b+6c &=31 \end{aligned} c + 1 4 b + 6 c = b = 3 1 B c + 1 = b 4 b + 6 c = 31 \begin{aligned} c+1 &= b \\ \\ 4b+6c &=31 \end{aligned} c + 1 4 b + 6 c = b = 3 1
- (Choice C) c − 1 = b 6 b + 4 c = 31 \begin{aligned} c-1 &= b \\ \\ 6b+4c &=31 \end{aligned} c − 1 6 b + 4 c = b = 3 1 C c − 1 = b 6 b + 4 c = 31 \begin{aligned} c-1 &= b \\ \\ 6b+4c &=31 \end{aligned} c − 1 6 b + 4 c = b = 3 1
- (Choice D) c + 1 = b 6 b + 4 c = 31 \begin{aligned} c+1 &= b \\ \\ 6b+4c &=31 \end{aligned} c + 1 6 b + 4 c = b = 3 1 D c + 1 = b 6 b + 4 c = 31 \begin{aligned} c+1 &= b \\ \\ 6b+4c &=31 \end{aligned} c + 1 6 b + 4 c = b = 3 1
- (Choice A) 1 1 1 1 A 1 1 1 1
- (Choice B) 2 2 2 2 B 2 2 2 2
- (Choice C) 3 3 3 3 C 3 3 3 3
- (Choice D) 4 4 4 4 D 4 4 4 4
- Your answer should be
- an integer, like 6 6 6 6
- a simplified proper fraction, like 3 / 5 3/5 3 / 5 3, slash, 5
- a simplified improper fraction, like 7 / 4 7/4 7 / 4 7, slash, 4
- a mixed number, like 1 3 / 4 1\ 3/4 1 3 / 4 1, space, 3, slash, 4
- an exact decimal, like 0.75 0.75 0 . 7 5 0, point, 75
- a multiple of pi, like 12 pi 12\ \text{pi} 1 2 pi 12, space, start text, p, i, end text or 2 / 3 pi 2/3\ \text{pi} 2 / 3 pi 2, slash, 3, space, start text, p, i, end text
Want to join the conversation?
- Upvote Button opens signup modal
- Downvote Button opens signup modal
- Flag Button opens signup modal
We’re Open - Call Now!
Writing Systems of Linear Equations from Word Problems
Some word problems require the use of systems of linear equations . Here are clues to know when a word problem requires you to write a system of linear equations:
(i) There are two different quantities involved: for instance, the number of adults and the number of children, the number of large boxes and the number of small boxes, etc. (ii) There is a value associated with each quantity: for instance, the price of an adult ticket or a children's ticket, or the number of items in a large box as opposed to a small box.
Such problems often require you to write two different linear equations in two variables. Typically, one equation will relate the number of quantities (people or boxes) and the other equation will relate the values (price of tickets or number of items in the boxes).
Here are some steps to follow:
1. Understand the problem.
Understand all the words used in stating the problem. Understand what you are asked to find. Familiarize the problem situation.
2. Translate the problem to an equation.
Assign a variable (or variables) to represent the unknown. Clearly state what the variable represents.
3. Carry out the plan and solve the problem.
Use substitution , elimination or graphing method to solve the problem.
The cost of admission to a popular music concert was $ 162 for 12 children and 3 adults. The admission was $ 122 for 8 children and 3 adults in another music concert. How much was the admission for each child and adult?
1 . Understand the problem:
The admission cost for 12 children and 3 adults was $ 162 . The admission cost for 8 children and 3 adults was $ 122 .
2 . Translate the problem to an equation.
Let x represent the admission cost for each child. Let y represent the admission cost for each adult. The admission cost for 12 children plus 3 adults is equal to $ 162 . That is, 12 x + 3 y = 162 . The admission cost for 8 children plus 3 adults is equal to $122. That is, 8 x + 3 y = 122 .
3 . Carry out the plan and solve the problem.
Subtract the second equation from the first. 12 x + 3 y = 162 8 x + 3 y = 122 _ 4 x = 40 x = 10 Substitute 10 for x in 8 x + 3 y = 122 . 8 ( 10 ) + 3 y = 122 80 + 3 y = 122 3 y = 42 y = 14 Therefore, the cost of admission for each child is $ 10 and each adult is $ 14 .
Download our free learning tools apps and test prep books
Names of standardized tests are owned by the trademark holders and are not affiliated with Varsity Tutors LLC.
4.9/5.0 Satisfaction Rating over the last 100,000 sessions. As of 4/27/18.
*See complete details for Better Score Guarantee.
Media outlet trademarks are owned by the respective media outlets and are not affiliated with Varsity Tutors.
Award-Winning claim based on CBS Local and Houston Press awards.
Varsity Tutors does not have affiliation with universities mentioned on its website.
Varsity Tutors connects learners with experts. Instructors are independent contractors who tailor their services to each client, using their own style, methods and materials.
IMAGES
VIDEO
COMMENTS
Section 2.2 : Linear Equations · 4x−7(2−x)=3x+2 4 x − 7 ( 2 − x ) = 3 x + 2 Solution · 2(w+3)−10=6(32−3w) 2 ( w + 3 ) − 10 = 6 ( 32 − 3 w )
Examples on Solving Linear Equations: · 1. Solve: (2x + 5)/(x + 4) = 1. Solution: (2x + 5)/(x + 4) = 1. ⇒ 2x + 5 = 1(x + 4) · 2. Solve: 6x - 19 = 3x - 10.
Word Problems on Linear Equations · Then the other number = x + 9. Let the number be x. · 2.The difference between the two numbers is 48. · 3. The length of a
Watch Sal work through a basic Linear equations word problem. ... You can solve a question any way you want as long as you're sure it's
Select variables to represent the unknown quantities. · Using the given information, write a system of two linear equations relating the two variables. · Solve
GRADE 7:SOLVING PROBLEMS INVOLVING LINEAR EQUATIONS IN ONE VARIABLEGRADE 7 PLAYLISTFirst Quarter: https://tinyurl.com/yyzdequa Second Qu...
Learn how to solve a word problem by writing an equation to model the situation. In this video, we use the linear equation 210(t-5) = 41,790
1.20: Word Problems for Linear Equations · 5x=60 · x · x=605=12 · 13. · 2x−5=13 · x · 2x=13+5, so that 2x=18 · x=182=9
1. Solve for x: 3x - 12 = 0. basket1aa · 2. Solve for m: 2(m + 6) = 48. purplegirl1 · 3. Solve for x: 3(2x - 1) - 10 = 8 + 5x. basket2 · 4. Solve for x: 8x + 9 -
Writing Systems of Linear Equations from Word Problems · 1. Understand the problem. Understand all the words used in stating the problem. Understand what you are